Term Rewriting System R:
[X, Y, X1, X2]
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ANATS -> AADX(azeros)
ANATS -> AZEROS
AADX(cons(X, Y)) -> AINCR(cons(X, adx(Y)))
AHD(cons(X, Y)) -> MARK(X)
ATL(cons(X, Y)) -> MARK(Y)
MARK(nats) -> ANATS
MARK(adx(X)) -> AADX(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(zeros) -> AZEROS
MARK(incr(X)) -> AINCR(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(tl(X)) -> ATL(mark(X))
MARK(tl(X)) -> MARK(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(tl(X)) -> MARK(X)
ATL(cons(X, Y)) -> MARK(Y)
MARK(tl(X)) -> ATL(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
AHD(cons(X, Y)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(tl(X)) -> MARK(X)
MARK(tl(X)) -> ATL(mark(X))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)
anats -> aadx(azeros)
anats -> nats
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  0  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(a__zeros)=  0  
  POL(incr(x1))=  x1  
  POL(A__TL(x1))=  x1  
  POL(a__hd(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(a__tl(x1))=  1 + x1  
  POL(tl(x1))=  1 + x1  
  POL(a__adx(x1))=  x1  
  POL(A__HD(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(hd(x1))=  x1  
  POL(nats)=  0  
  POL(s(x1))=  0  
  POL(zeros)=  0  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

ATL(cons(X, Y)) -> MARK(Y)
MARK(hd(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
AHD(cons(X, Y)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(hd(X)) -> MARK(X)
AHD(cons(X, Y)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(hd(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)
anats -> aadx(azeros)
anats -> nats
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(a__nats)=  0  
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(a__zeros)=  0  
  POL(incr(x1))=  x1  
  POL(a__hd(x1))=  1 + x1  
  POL(mark(x1))=  x1  
  POL(a__tl(x1))=  x1  
  POL(tl(x1))=  x1  
  POL(a__adx(x1))=  x1  
  POL(A__HD(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(hd(x1))=  1 + x1  
  POL(nats)=  0  
  POL(s(x1))=  0  
  POL(zeros)=  0  
  POL(a__incr(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pairs:

AHD(cons(X, Y)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(incr(X)) -> MARK(X)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  x1  
  POL(incr(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pair:

MARK(adx(X)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




The following dependency pair can be strictly oriented:

MARK(adx(X)) -> MARK(X)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(adx(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes